Chapter 4.5, Problem 4E

a.

To determine

To find the linearization $L(x)$ of $f(x)=\ln (x+1)$ at $x=a$ and $a=0$ .

Answer to Problem 4E

Linearization $L(x)$ of $f(x)=\ln (x+1)$ at $a=0$ is $L(x)=x$.

Explanation of Solution

Given: The function $f(x)=\ln (x+1)$ at $a=0$

Formula used: $L(x)=f(a)+f\text{'}(a)(x-a)$

Calculation:

The given function is $f(x)=\ln (x+1)$ .

The formula used to find linearization of a function is $L(x)=f(a)+f\text{'}(a)(x-a)$ .

  $f(a)=\ln (a+1)$

Substitute $a=0$ in $f(a)$

  $\begin{array}{l}f(a)=\ln (0+1)\\ f(0)=\ln 1\\ f(0)=0\end{array}$

Differentiate $f(a)$ with respect to a

  $f\text{'}(a)=\frac{1}{(x+1)}$

Substitute $a=1$ in $f\text{'}(x)$

  $\begin{array}{l}f\text{'}(a)=\frac{1}{(0+1)}\\ f\text{'}(1)=1\end{array}$

Substitute $f(a)$ and $f\text{'}(x)$ in the formula

  $\begin{array}{l}L(x)=0+1(x-0)\\ L(x)=x\end{array}$

Conclusion:

Linearization $L(x)$ of $f(x)=\ln (x+1)$ at $a=1$ is $L(x)=x$.

b.

To determine

The accuracy of approximation of $L(a+0.1)=f(a+0.1)$

Answer to Problem 4E

  $L(a+0.1)\simeq f(a+0.1)$ is more approximate.

Explanation of Solution

Given:

The function is $f(a)=\ln (a+1)$ .

The linearization function is $L(x)=x$ .

  $\begin{array}{l}f(a+0.1)=\ln (a+0.1+1)\\ L(a+0.1)=a+0.1\end{array}$

Substituting $a=0$ in above equations

  $\begin{array}{l}f(a+0.1)=\ln (0+0.1+1)\\ f(a+0.1)=0.0953\\ L(a+0.1)=0+0.1\\ L(a+0.1)=0.1\end{array}$

It is more accurate.

Conclusion:

  $L(a+0.1)\simeq f(a+0.1)$ is more approximate.